Indoor Air Unit Conversion Background - US EPA

Indoor Air Unit Conversion

Background

In dilute aqueous systems at room temperature and 1 atmosphere of pressure, 1 liter (L) of water weighs

1 kilogram (kg). Therefore, 1 milligram (mg) of a contaminant in 1 liter (L) of water has a concentration

of 1 mg/L, which is the same as 1 mg of containment/1 kg of water on a mass/mass basis. Since there

are 1 million mg in 1 kg, the kg in the denominator may be converted to 1 million mg. So our 1 mg/L

solution is equivalent to 1 mg/1,000,000 mg. This is referred to as ¡°1 part per million¡± or ppm in aqueous

solutions. Similarly, 1?g/L is referred to as ¡°1 part per billion¡± or ppb in dilute aqueous solutions because

there are 1 billion micrograms in 1 kg.

However, indoor air units are not expressed as a mass-per-mass ratio, even though they are given as

ppm or ppb. The units of ppm and ppb in gas systems are computed on a volume-per-volume ratio and

should more accurately be termed ppmV and ppbV. For example:

So, how do we convert between the mass-per-volume units and ppmV or ppbV in a gas system?

?

First, we must use the ideal gas law to convert the measured contaminant mass to a volume.

The ideal gas law (PV=nRT) relates pressure, volume, temperature and mass of a gaseous

contaminant:

1.

where P air is air pressure

V contaminant is the volume occupied by the contaminant

R is the universal gas constant, and

T air is air temperature. (¡° ¡± represents multiplication.)

Any units for pressure, volume and temperature may be used, as long as the universal gas

constant is in consistent units. Noting that # moles contaminant =mass contaminant /molecular

weight contaminant , and using pressure, temperature and volume in units of [kPa], [K[ and [L], we can

solve the preceding relationship for the volume of our contaminant, given its mass in grams:

2.

Note that T[K] = T[?C] + 273.15.

?

Now that we have the mass of the contaminant converted to a volume, we simply need to divide

by the volume of the sample measurement, and work out the units. For example, ppmV is

equivalent to 1 mL/m3 and ppbV is equivalent to 1 ?L/m3. Or in equation form:

3.

?

and

So, to convert from ?g/m3 to ppmV, we plug in our mass values in equation 2 above, making sure

to convert our ?g to units of grams required by the equation. This will give us the volume of our

contaminant in liters. We must now convert this into mL for equation 3. Then we simply divide

by the sample volume in m3 to obtain our result in ppmV. Likewise, to convert ?g/m3 to ppbV,

we would follow the same procedure, except we¡¯d convert the volume of the contaminant to ?L

instead of mL.

Example

For a numerical example, let¡¯s convert 123.45 ?g/m3 of benzene to ppmV. We¡¯ll assume 25 ?C and 1

atmosphere pressure (101.325 kPa). So using equation 2, 123.45 ?g (which is 123.45 x 10-6 grams) of

benzene (which has a molecular weight of 78.11 g/mole) occupies the following volume:

= 3.866 x 10-5 L or 0.03866 mL.

Dividing this by the sample volume in m3(=1 m3) gives us our result in ppmV:

123.45 ?g/m3 of benzene at 25 ?C and 1 atm pressure = 0.0386 ppmV.

For more information, see Introduction to Air Toxics Analyses by Don Harrington of Teledyne

instruments.

Here are the conversions used in the online calculator, all based on a equations 2 and 3 and appropriate

units:

?g/m3 to ppmV

mg/m3 to ppmV

?g/L to ppmV

mg/L to ppmV

?g/m3 to ppbV

mg/m3 to ppbV

?g/L to ppbV

mg/L to ppbV

Here are some other useful conversions:

ppmV x 1,000 = ppbV

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